Question

For the reaction at 298 K : 2A + B → C 

∆H = 400 KJ mol^-1: ∆S = 0.2 JK^-1 mol^-1 

Determine the temperature at which the reaction would be spontaneous.

Answer 1

Data given, 

2A + B → C at 298 K 

∆H = 400 KJ mol^-1 

∆S = 0.2 JK^-1 mol^-1 

T = 298 K 

[∆G = ∆H – T∆S] 

if ∆G = 0 

∆H – T∆S = 0

∆S = ∆H / T

For ∆G = 0, 

∆S = ∆H / T 

= 400 / 0.2

= 4000 / 2 

= 2000 K

At 2000 K. the reaction is in equilibrium. So. above 2000 K, the reaction will be spontaneous.